# Thread: negative acceleration ( confusion!!!!!!!!!!!)

1. A pachutist at the instant of first contact with th ground has two forces acting on him contact push and the gravititional force but the cantact push force is bigger . I found the reason for that but i did not understand it. It says the parachutist has a negative acceleration which requires a net upward force!!!!!!!!!!
can u explain it for me???????

2.

3. Originally Posted by almirza
A pachutist at the instant of first contact with th ground has two forces acting on him contact push and the gravititional force but the cantact push force is bigger . I found the reason for that but i did not understand it. It says the parachutist has a negative acceleration which requires a net upward force!!!!!!!!!!
can u explain it for me???????
He has to stop. Prior to the impact, he has a velocity v (i.e. a momentum p towards the earth). After the impact, he has a velocity of 0 relative to the earth, and hence a momentum of 0 relative to the earth. Since F = p-dot and since p-dot is away from the earth, the contact force during the impact must on average exceed the force of gravity.

Always remember that velocity and acceleration are vector quantities. While the magnitude will always be positive a vector also has a direction, and in one-dimensional problems it is common to conflate the vector and scalar notation and denote direction with the sign (i.e. -v means that the object is moving at a speed of v in the minus direction).

Although in the usual approximation of a wholly inelastic, instantaneous collision between incompressible objects, the contact force becomes infinite (which is not a problem, because it only does work in an infinitisimal interval of time), so I don't really see point in the exercise.

4. I sure hope that he was "deployed". :wink:

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